Question

Sharon is clicking a picture of her grandmother who is standing 5.4 m in front of a camera, which has a converging lens with a focal length of 0.075 m. The distance of the image from the lens is

Answer 1

Explanation:

general lens formula,

1/f = 1/v - 1/u

for converging lens,

f = + ve and u = - ve

1/f = 1/ v + 1/5.4

1/ 0.075 = 1/5.4 +1/v

1/v = 1/0.075 - 1/5.4

=( 5.4 - 0.075)/ 0.075 x 5.4

= 5.325 /0.4050

v = 0.4050 / 5.325 = 0.076m

Answer 2

Answer:

The distance of the image from the lens is 0.076m.

Explanation:

From the lens formula we have,

$\frac{1}{f} =\frac{1}{v} -\frac{1}{u}$            (1)

Where,

f=focal length of the lens

v=image distance from the lens

u=object distance from the lens

From the question we have,

f=+0.075m    ("+" as the lens is converging)

u=-5.4m        ("-" as the object is real)

By substituting the values in equation (1) we get;

$\frac{1}{0.075} =\frac{1}{v} -\frac{1}{-5.4}$

$\frac{1}{0.075} =\frac{1}{v} +\frac{1}{5.4}$

$\frac{1}{v}=\frac{1}{0.075}-\frac{1}{5.4}$

$\frac{1}{v}=\frac{5.4-0.075}{0.075\times 5.4}$

$\frac{1}{v}=\frac{5.325}{0.4050}$

$v=\frac{0.4050}{5.325}$

$v=0.076m$

Hence, the distance of the image from the lens is 0.076m.